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FirstList = {{1, 2, x}, {2, 3, x}, {4, 2, x}, {x, 3, 5}, {x, 7, 8}};
SecondList = {a, b, c, d, e};
Table[
Replace[Part[FirstList, n], x -> Part[SecondList, n]], {n, Length[SecondList]}]
{{1, 2, x}, {2, 3, x}, {4, 2, x}, {x, 3, 5}, {x, 7, 8}}
I expected the output to be
{{1, 2, a}, {2, 3, b}, {4, 2, c}, {d, 3, 5}, {e, 7, 8}}
How do you make it so?
The level specification of Replace is wrong. Compare:
Replace[x, x -> a]
Replace[{1, 2, x}, x -> a]
Replace[{1, 2, x}, x -> a, {1}]
Out[14]= a
Out[15]= {1, 2, x}
Out[16]= {1, 2, a}
So to make this work, add a level specification to Replace:
Table[Replace[Part[FirstList, n], x -> Part[SecondList, n], {1}], {n, Length[SecondList]}]
You can also use ReplaceAll, which operates at every level. In general, though, I prefer Replace where possible over the scattershot approach of ReplaceAll.
ReplaceAll (operator form: /.) automatically moves through all levels, so in this case it will do its replacements at level 1 and give the result you want.
a1 = {{1, 2, x}, {2, 3, x}, {4, 2, x}, {x, 3, 5}, {x, 7, 8}};
a2 = {a, b, c, d, e};
MapThread[(#1 /. x -> #2) &, {a1, a2}]
{{1, 2, a}, {2, 3, b}, {4, 2, c}, {d, 3, 5}, {e, 7, 8}}
ReplaceAll works at level 1 by default", because ReplaceAll does not have a level spec and works on every level (not just level 1).
$\endgroup$
– Sjoerd Smit
Mar 7 '18 at 14:01
ReplaceAll also works at level 0... Or am I confusing level specs with positions here?
$\endgroup$
– Sjoerd Smit
Mar 7 '18 at 14:15
Replace[x, x -> 1, {0}] with ReplaceAll[x, x -> 1]. Both do the replacement, which happens at level 0.
$\endgroup$
– Sjoerd Smit
Mar 7 '18 at 14:19
